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\title{\heiti\zihao{2}习题13.1}
\author{中书君}
\date{\today}

\begin{document}
\maketitle
\section{证明: 勒让德(Legendre)多项式$$\left\{\begin{array}{l}p_{0}(x)=1 \\p_{n}(x)=\dfrac{1}{2^{n} n !} \dfrac{\mathrm{d}^{n}\left(x^{2}-1\right)^{n}}{\mathrm{~d} x^{n}}, n \in \mathrm{N}^{*}\end{array}\right.$$是$[-1,1]$上的正交函数系.}
\begin{proof}
	等价于证明
	$$
		\int_{-1}^{1} p_{n}(x) p_{m}(x) \mathrm{~d} x\left\{\begin{array}{ll}
			=0,     & n \neq m, \\
			\neq 0, & n=m,
		\end{array}\right.
	$$
	将其内部进行代换,可得:
	$$
		\begin{aligned}
			\int_{-1}^{1} p_{n}(x) p_{m}(x) \mathrm{~d} x & = \int_{-1}^{1}\dfrac{1}{2^{n}n!}\dfrac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n}\dfrac{1}{2^{m}m!}\dfrac{\mathrm{d}^{m}}{\mathrm{d}x^{m}}(x^{2}-1)^{m}\mathrm{~d}x \\
			                                              & =\dfrac{1}{2^{m+n}}\dfrac{1}{m!n!}\int_{-1}^{1}\dfrac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{m}}{\mathrm{d}x^{m}}(x^{2}-1)^{m}\mathrm{~d}x
		\end{aligned}
	$$
	而当$m\neq n $时,要证明$\int_{-1}^{1} p_{n}(x) p_{m}(x) \mathrm{~d} x=0$只需要$\int_{-1}^{1}\dfrac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{m}}{\mathrm{d}x^{m}}(x^{2}-1)^{m}\mathrm{~d}x=0$.
	对其进行分部积分:
	$$
		\begin{aligned}
			\int_{-1}^{1}\dfrac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{m}}{\mathrm{d}x^{m}}(x^{2}-1)^{m}\mathrm{~d}x  = & \left.\dfrac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{m}}{\mathrm{d}x^{m}}(x^{2}-1)^{m}\right|_{-1}^{1}         \\
			                                                                                                                                     & -\int_{-1}^{1}\dfrac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{m+1}}{\mathrm{d}x^{m+1}}(x^{2}-1)^{m}\mathrm{~d}x
		\end{aligned}
	$$
	由于$\dfrac{\mathrm{d^{n-1}}}{\mathrm{d}x^{n-1}}(x^{2}-1)^{n}=u'(x)\cdot \left[(u(x))^{n}\right]^{(n-1)}=u'(x)\cdot n! u(x)=g(x)$,其中$u(x)=(x^{2}-1)$.由于$u(x)\mid g(x)$,所以$x=\pm 1$时$g(x)=0$,从而
	$$
		\begin{aligned}
			\int_{-1}^{1}\dfrac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{m}}{\mathrm{d}x^{m}}(x^{2}-1)^{m}\mathrm{~d}x & = -\int_{-1}^{1}\dfrac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{m+1}}{\mathrm{d}x^{m+1}}(x^{2}-1)^{m}\mathrm{~d}x     \\
			                                                                                                                                  & =(-1)^{m}\int_{-1}^{1}\dfrac{\mathrm{d}^{n-m}}{\mathrm{d}x^{n-m}}(x^{2}-1)^{n}\dfrac{\mathrm{d}^{2m}}{\mathrm{d}x^{2m}}(x^{2}-1)^{m}\mathrm{~d}x \\
			                                                                                                                                  & =(-1)^{m}2m!\int_{-1}^{1}\dfrac{\mathrm{d}^{n-m}}{\mathrm{d}x^{n-m}}(x^{2}-1)^{n}\mathrm{~d}x                                                    \\
			                                                                                                                                  & =(-1)^{m}2m!\left.\dfrac{\mathrm{d}^{n-m-1}}{\mathrm{d}x^{n-m-1}}(x^{2}-1)^{n}\right|_{-1}^{1}
		\end{aligned}
	$$
	同上理,积分表达式中有$u(x)=x^{2}-1$多项式因子,所以积分为$0$.\par
	当$m=n$时,计算
	$$
		\begin{aligned}
			\dfrac{1}{2^{2n}}\dfrac{1}{(n!)^{2}}\int_{-1}^{1}(x^{2}-1)^{n}\mathrm{~d}x & =\dfrac{1}{2^{2n}}\dfrac{1}{(n!)^{2}}\int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\cos^{2n+1}u\mathrm{~d}u \\
			                                                                           & =\dfrac{1}{2^{2n}}\dfrac{1}{(n!)^{2}}\dfrac{2\cdot (2n)!!}{(2n+1)!!}(2n)!                            \\
			                                                                           & =\dfrac{2}{2n+1}
		\end{aligned}
	$$
	而显然当$m=n=0$时积分为常值积分,所以值为$2$,满足$\dfrac{2}{2n+1}$\par
	所以
	$$
		\int_{-1}^{1}p_{n}(x)p_{m}(x)=\left\{\begin{array}{ll}
			0,               & n \neq m \\
			\dfrac{2}{2n+1}, & n=m
		\end{array}\right.
	$$
	从而其为$[-1,1]$上的正交函数系.
\end{proof}

\section{求下列函数的 Fourier 级数}
\subsection{$f(x)=\left\{\begin{array}{ll}x, & 0\leqslant x \leqslant \pi \\ 0, & -\pi<x<0 ;\end{array}\right.$}
\textbf{解}\quad
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\mathrm{~d}x=\dfrac{\pi}{2}                                   \\
		a_{n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\mathrm{~d}x                                           \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}x\cos nx \mathrm{~d}x                                                \\
		      & =\dfrac{1}{n\pi}\left.x\sin nx\right|_{0}^{\pi}-\dfrac{1}{n\pi}\int_{0}^{\pi}\sin nx\mathrm{~d}x  \\
		      & =\dfrac{1}{n^{2}\pi}\left.\cos nx\right|_{0}^{\pi}                                                \\
		      & =\dfrac{(-1)^{n}-1}{n^{2}\pi}                                                                     \\
		b_{n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx\mathrm{~d}x                                           \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}x\sin nx\mathrm{~d}x                                                 \\
		      & =-\dfrac{1}{n\pi}\left.x\cos nx\right|_{0}^{\pi}+\dfrac{1}{n\pi}\int_{0}^{\pi}\cos nx\mathrm{~d}x \\
		      & =\dfrac{(-1)^{n+1}}{n}
	\end{aligned}
$$
由于$f(x)$只有第一类间断点,而且在间断点附近存在左右单侧导数,所以
$$
	f(x)\sim\dfrac{\pi}{4}+\sum\limits_{n=1}^{\infty}\left(\left(-\dfrac{1}{2}+\dfrac{(-1)^{n}}{2}\right)\dfrac{2}{n^{2}\pi}\cos nx+\dfrac{(-1)^{n+1}}{n}\sin nx\right)$$
并且
$$
	\dfrac{\pi}{4}+\sum\limits_{n=1}^{\infty}\left(\left(-\dfrac{1}{2}+\dfrac{(-1)^{n}}{2}\right)\dfrac{2}{n^{2}\pi}\cos nx+\dfrac{(-1)^{n+1}}{n}\sin nx\right)=\dfrac{f(x-0)+f(x+0)}{2}
$$
\subsection{$f(x)=x \sin x, \quad x \in[-\pi, \pi]$}
\textbf{解}
$f(x)$为偶函数,从而$b_{n}=0$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}x\sin x\mathrm{~d}x = \dfrac{2}{\pi}\int_{0}^{\pi}x\sin x\mathrm{~d}x \\
		      & =\dfrac{2}{\pi}\left.x\cos x\right|_{0}^{\pi}+\dfrac{2}{\pi}\int_{0}^{\pi}\cos x\mathrm{~d}x          \\
		      & =2                                                                                                    \\
		a_{1} & =\dfrac{2}{\pi}\int_{0}^{\pi}x\sin x\cos x                                                            \\
		      & =-\dfrac{1}{2\pi}\left.x\cos 2x\right|_{0}^{\pi}-\int_{0}^{\pi}\cos 2x\mathrm{~d}x=-\dfrac{1}{2}      \\
		a_{n} & =\dfrac{2}{\pi}\int_{0}^{\pi}x\sin x \cos nx\mathrm{~d}x                                              \\
		      & =\dfrac{2(-1)^{n}}{1-n^{2}}
	\end{aligned}
$$
$f(x)$无间断点,且在定义域内可微,从而
$$
	f(x)=1-\dfrac{1}{2}\cos x+2\sum_{n=2}^{\infty}\dfrac{(-1)^{n}}{1-n^{2}}
$$

\subsection{$f(x)=\left\{\begin{array}{ll}x^{2}, & 0<x<\pi \\ 0, & x=\pi \\ -x^{2}, & \pi<x \leqslant 2 \pi\end{array}\right.$}
\textbf{解}
$f(x)$不是奇函数也不是偶函数.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\mathrm{~d}x                                                               \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}x^{2}\mathrm{~d}x-\dfrac{1}{\pi}\int_{\pi}^{2\pi}x^{2}\mathrm{~d}x                \\
		      & =-2\pi^{2}                                                                                                     \\
		a_{n} & =\dfrac{1}{\pi}\int_{0}^{2\pi}f(x)\cos nx\mathrm{~d}x                                                          \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}x^{2}\cos nx\mathrm{~d}x-\dfrac{1}{\pi}\int_{\pi}^{2\pi}x^{2}\cos nx\mathrm{~d}x
		\\
		      & =\dfrac{4((-1)^{n}-1)}{n^{2}}                                                                                  \\
		b_{n} & =\dfrac{1}{\pi}\int_{0}^{2\pi}f(x)\sin nx \mathrm{~d}x                                                         \\
		      & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}x^{2}\sin nx\mathrm{~d}x -\int_{\pi}^{2\pi}x^{2}\sin nx \mathrm{~d}x\right) \\
		      & =-\dfrac{4\pi}{n}-\dfrac{2\pi(-1)^{n}}{n}+\dfrac{4((-1)^{n}-1)}{\pi n^{3}}
	\end{aligned}
$$
所以$f(x)\sim-\pi^{2}+\sum\limits_{n=1}^{\infty}\left(\dfrac{4((-1)^{n}-1)}{n^{2}}\cos nx+\left(-\dfrac{4\pi}{n}-\dfrac{2\pi(-1)^{n}}{n}+\dfrac{4((-1)^{n}-1)}{\pi n^{3}}\right)\sin nx\right)$,并且由于$f(x)$分段可微(分段光滑),所以在间断点处级数收敛于$\dfrac{f(x-0)+f(x+0)}{2}$.
\subsection{$f(x)=|x|, \quad x \in[-\pi, \pi] $}
\textbf{解}
$f(x)$是偶函数,$b_{n}=0$.其无间断点,且在定义域内分段可微,从而
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\mathrm{~d}x                                                     \\
		      & =\pi                                                                                                 \\
		a_{n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx \mathrm{~d}x                                             \\
		      & =-\dfrac{1}{\pi}\int_{-\pi}^{0}x\cos nx\mathrm{~d}x+\dfrac{1}{\pi}\int_{0}^{\pi}x\cos nx\mathrm{~d}x \\
		      & =\dfrac{2(-1)^{n}-2}{n^{2}}
	\end{aligned}
$$
所以$f(x)=\dfrac{\pi}{2}+\sum\limits_{n=1}^{\infty}\left(\dfrac{2(-1)^{n}-2}{n^{2}}\cos nx\right)$.

\subsection{$f(x)=\left\{\begin{array}{cl}0, & -\pi \leqslant x<0, \\ \sin x, & 0 \leqslant x<\pi ;\end{array}\right.$}
\textbf{解}
$f(x)$无间断点且分段可微,从而
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\mathrm{~d}x                \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}\sin x\mathrm{~d}x=\dfrac{2}{\pi}  \\
		a_{1} & =\dfrac{1}{\pi}\int_{0}^{\pi}\sin x\cos x\mathrm{~d}x=0         \\
		a_{n} & =\dfrac{1}{\pi}\int_{0}^{\pi}\sin x \cos nx\mathrm{~d}x         \\
		      & =\dfrac{1+(-1)^{n}}{\pi(n^{2}-1)}                               \\
		b_{1} & =\dfrac{1}{\pi}\int_{0}^{\pi}\sin^{2}x\mathrm{~d}x=\dfrac{1}{2} \\
		b_{n} & =\dfrac{1}{\pi}\int_{0}^{\pi}\sin x\sin nx\mathrm{~d}x          \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}\dfrac{-\cos(x+nx)+\cos(x-nx)}{2}  \\
		      & =\dfrac{2}{\pi}\int_{0}^{\pi}\cos (x-nx)\mathrm{~d}x            \\
		      & =\dfrac{\sin (1-n)\pi}{2(1-n)}                                  \\
		      & =0
	\end{aligned}
$$
所以$f(x)=\dfrac{1}{\pi}+\dfrac{1}{2}\sin x+\sum\limits_{n=2}^{\infty}\left(\dfrac{1+(-1)^{n}}{\pi(n^{2}-1)}\cos nx\right)$
\subsection{$f(x)=x \cos x, \quad x \in(-\pi, \pi) .$}
\textbf{解}
$f(x)$是奇函数所以$a_{n}=0$.
$$
	\begin{aligned}
		b_{1} & =\dfrac{2}{\pi}\int_{0}^{\pi}x\cos x \sin x\mathrm{~d}x                                                                                                                                                   \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}x\sin 2x\mathrm{~d}x                                                                                                                                                         \\
		      & =-\dfrac{1}{2}                                                                                                                                                                                            \\
		b_{n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}x\cos x\sin nx                                                                                                                                                            \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}x\left(\sin(n+1)x+\sin (n-1)x\right)\mathrm{~d}x                                                                                                                             \\
		      & =\left.-\dfrac{1}{\pi}x\left[\dfrac{\cos(n+1)x}{n+1}+\dfrac{\cos(n-1)x}{n-1}\right]\right|_{0}^{\pi}+\dfrac{1}{\pi}\int_{0}^{\pi}\left[\dfrac{\cos(n+1)x}{n+1}+\dfrac{\cos(n-1)x}{n-1}\right]\mathrm{~d}x \\
		      & =\left(\dfrac{(-1)^{n}}{n+1}+\dfrac{(-1)^{n}}{n-1}\right)+0                                                                                                                                               \\
		      & =\dfrac{2n(-1)^{n}}{n^{2}-1}
	\end{aligned}
$$
从而$f(x)\sim-\dfrac{1}{2}\sin x+2\sum\limits_{n=2}^{\infty}\dfrac{(-1)^{n}n}{n^{2}-1}\sin nx$.在间断点$\pm 2k+1\pi$处级数收敛到$\dfrac{f(x-0)+f(x+0)}{2}=0$.

\section{设 $f(x)$ 是周期为 $2 \pi$ 的可积或绝对可积函数,证明:}
\subsection{如果 $f(x)$ 在 $[-\pi, \pi]$ 中满足 $f(x+\pi)=f(x)$,那么 $f(x)$ 的 Fourier 级数系数满足 $a_{2 n-1}=b_{2 n-1}=0, \quad n \in \mathbb{N}^{*}$}
\begin{proof}
	$$
		\begin{aligned}
			a_{2n-1} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos (2n-1)x\mathrm{~d}x                                                          \\
			         & =\dfrac{1}{\pi}\int_{-\pi}^{0}f(x)\cos (2n-1)x\mathrm{~d}x+\dfrac{1}{\pi}\int_{0}^{\pi}f(x)\cos (2n-1)x\mathrm{~d}x   \\
			         & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}-f(x+\pi)\cos (2n-1)x\mathrm{~d}x+\int_{0}^{\pi}f(x)\cos(2n-1)x\mathrm{~d}x\right) \\
			         & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}\left(-f(x)+f(x)\right)\cos(2n-1)x\mathrm{~d}x\right)                              \\
			         & =0                                                                                                                    \\
			b_{2n-1} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(2n-1)x\mathrm{~d}x                                                           \\
			         & =\dfrac{1}{\pi}\int_{-\pi}^{0}f(x)\sin(2n-1)x\mathrm{~d}x+\dfrac{1}{\pi}\int_{0}^{\pi}f(x)\sin(2n-1)x\mathrm{~d}x     \\
			         & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}-f(x+\pi)\sin(2n-1)x\mathrm{~d}x+\int_{0}^{\pi}f(x)\sin(2n-1)x\mathrm{~d}x\right)  \\
			         & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}\left(-f(x)+f(x)\right)\sin(2n+1)x\mathrm{~d}x\right)                              \\
			         & =0
		\end{aligned}
	$$
\end{proof}

\subsection{如果 $f(x)$ 在 $[-\pi, \pi]$ 中满足 $f(x+\pi)=-f(x)$,那么 $f(x)$ 的 Fourier 级数系数 满足$$a_{2 n}=b_{2 n}=0, \quad n \in \mathbb{N}^{*}$$}
\begin{proof}
	$$
		\begin{aligned}
			a_{2n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos 2nx\mathrm{~d}x                                                      \\
			       & =\dfrac{1}{\pi}\int_{-\pi}^{0}f(x)\cos 2nx\mathrm{~d}x+\dfrac{1}{\pi}\int_{0}^{\pi}f(x)\cos 2nx\mathrm{~d}x   \\
			       & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}\left(-f(x)+f(x)\right)\cos 2nx\mathrm{~d}x\right)                         \\
			       & =0                                                                                                            \\
			b_{2n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin 2nx\mathrm{~d}x                                                      \\
			       & =\dfrac{1}{\pi}\int_{-\pi}^{0}f(x)\sin 2nx \mathrm{~d}x+\dfrac{1}{\pi}\int_{0}^{\pi}f(x)\sin 2nx \mathrm{~d}x \\
			       & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}\left(-f(x)+f(x)\right)\sin 2nx mathrm{~d}x\right)                         \\
			       & =0
		\end{aligned}
	$$
\end{proof}

\section{求下列函数的 Fourier 级数 }
\subsection{将函数 $f(x)=\cos \dfrac{x}{2}$ 在 $[0, \pi]$ 上展开为正弦级数}
\textbf{解}\quad
对函数做奇延拓,$a_{n}=0$.
$$
	\begin{aligned}
		b_{n} & =\dfrac{2}{\pi}\int_{0}^{\pi}\cos\dfrac{x}{2}\sin nx\mathrm{~d}x                                                        \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}\left[\sin\left(n+\dfrac{1}{2}\right)x+\sin\left(n-\dfrac{1}{2}\right)x\right]\mathrm{~d}x \\
		      & =\dfrac{8n}{(4n^{2}-1)\pi}
	\end{aligned}
$$
再由$f(x)$分段可微,从而在$(0,\pi]$上,$(x)=\sum\limits_{n=1}^{\infty}\dfrac{8n}{(4n^{2}-1)\pi}\sin nx$,在$0$处级数收敛到$$\dfrac{f(x-0)+f(x+0)}{2}=0$$

\subsection{将函数 $f(x)=\left\{\begin{array}{ll}1-x, & 0<x \leqslant 2, \\ x-3, & 2<x<4,\end{array}\right.$ 在$(0,4)$上展开为余弦级数}
\textbf{解}\quad
将$f(x)$偶延拓,$2T=4$,从而$b_{n}=0$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{T}\int_{0}^{2T}f(x)\mathrm{~d}x=0                    \\
		a_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\cos \dfrac{n\pi}{T}x\mathrm{~d}x \\
		      & =\dfrac{2}{T}\int_{0}^{T}(1-x)\cos \dfrac{n\pi}{T}x\mathrm{~d}x \\
		      & =-\dfrac{4((-1)^{n}-1)}{\pi^{2}n^{2}}
	\end{aligned}
$$
所以由$f(x)$分段可微且没有间断点可得
$$
	f(x)=-\sum_{n=1}^{\infty}\left(\dfrac{4((-1)^{n}-1)}{\pi^{2}n^{2}}\cos \dfrac{n\pi}{2}x\right)
$$
\subsection{将函数 $f(x)=\dfrac{\pi}{2}-x$ 在 $[0, \pi]$ 上展开为余弦级数}
将$f(x)$偶延拓,$b_{n}=0$,$2T=2\pi.$
\textbf{解}\quad
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\mathrm{~d}x=0                  \\
		a_{n} & =\dfrac{2}{\pi}\int_{0}^{\pi}(\dfrac{\pi}{2}-x)\cos n x\mathrm{~d}x \\
		      & =\dfrac{2(1-(-1)^{n})}{\pi n^{2}}
	\end{aligned}
$$
由$f(x)$分段可微且无间断点知其在$[0,\pi]$上展开为
$$
	f(x)=\sum_{n=1}^{\infty}\left(\dfrac{2(1-(-1)^{n})}{\pi n^{2}}\cos nx\right)
$$
\subsection{将函数 $f(x)=1+x$ 在 $(0, \pi)$ 上分别展开为正弦级数和余弦级数}
\textbf{解}\par
(1)对$f(x)$进行偶延拓,从而$b_{n}=0$,$2T=2\pi$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\mathrm{~d}x                                                               \\
		      & =\pi+2                                                                                                         \\
		a_{n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\mathrm{~d}x                                                        \\
		      & =\dfrac{2}{\pi}\int_{0}^{\pi}f(x)\cos nx\mathrm{~d}x     =\dfrac{2}{\pi}\int_{0}^{\pi}(1+x)\cos nx\mathrm{~d}x \\
		      & =\dfrac{2((-1)^{n}-1)}{n^{2}\pi}
	\end{aligned}
$$
$f(x)$分段可微,所以$f(x)=1+\dfrac{\pi}{2}+\sum\limits_{n=1}^{\infty}\left(\dfrac{2((-1)^{n}-1)}{n^{2}\pi}\cos nx\right)$.\par
(2)对$f(x)$进奇延拓,则$a_{n}=0$.
$$\
	\begin{aligned}
		b_{n} & =\dfrac{2}{\pi}(x+1)\sin nx\mathrm{~d}x \\
		      & =\dfrac{2(1-(-1)^{n}(\pi+1))}{n\pi}
	\end{aligned}
$$
$f(x)$分段可微,所以$f(x)=\sum\limits_{n=1}^{\infty}\left(\dfrac{2(1-(-1)^{n}(\pi+1))}{n\pi}\sin nx\right)$.

\section{求下列函数的 Fourier 级数:}
\subsection{$f(x)=\left\{\begin{array}{ll}0, & -5 \leqslant x<0, \\ 3, & 0 \leqslant x<5 ;\end{array}\right.$}
\textbf{解}\quad
将$f(x)$延拓,得到周期为$10$的函数.记$2T=10$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\mathrm{~d}x                      \\
		      & =\dfrac{1}{5}\int_{0}^{5}3\mathrm{~d}x=3                        \\
		a_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\cos \dfrac{n\pi}{T}x\mathrm{~d}x \\
		      & =\dfrac{1}{5}\int_{0}^{5}3\cos \dfrac{n\pi}{5}x\mathrm{~d}x=0   \\
		b_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\sin \dfrac{n\pi}{T}x\mathrm{~d}x \\
		      & =\dfrac{1}{5}\int_{0}^{5}3\sin\dfrac{n\pi}{5}x\mathrm{~d}x      \\
		      & =\dfrac{3(1-(-1)^{n})}{n\pi}
	\end{aligned}
$$
所以由于$f(x)$分段可微,从而$f(x)=\dfrac{3}{2}+\sum\limits_{n=1}^{\infty}\left(\dfrac{3(1-(-1)^{n})}{n\pi}\sin \dfrac{n\pi}{5}x\right)$.间断点处级数收敛于$\dfrac{f(x-0)+f(x+0)}{2}=\dfrac{3}{2}$.

\subsection{$f(x)=\left\{\begin{array}{ll}x, & 0 \leqslant x \leqslant 1, \\ 1, & 1<x<2 \\ 3-x, & 2\leqslant x \leqslant 3\end{array}\right.$}
\textbf{解}\quad
将$f(x)$延拓,其为周期为$3$的偶函数,所以$b_{n}=0$,记$2T=3$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\mathrm{~d}x=\dfrac{4}{3}                                                                            \\
		a_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\cos \dfrac{n\pi}{T}x\mathrm{~d}x                                                                    \\
		      & =\dfrac{4}{3}\int_{0}^{\frac{3}{2}}f(x)\cos \dfrac{2n\pi}{3}x\mathrm{~d}x                                                          \\
		      & =\dfrac{4}{3}\left(\int_{0}^{1}x\cos \dfrac{2n\pi}{3}x\mathrm{~d}x+\int_{1}^{\frac{3}{2}}\cos \dfrac{2n\pi}{3}x\mathrm{~d}x\right) \\
		      & =\dfrac{3\left(\cos\dfrac{2\pi n}{3}-1\right)}{\pi^{2}n^{2}}
	\end{aligned}
$$
所以由于$f(x)$分段可微,从而$f(x)=\dfrac{2}{3}+\sum\limits_{n=1}^{\infty}\left(\dfrac{3\left(\cos\dfrac{2\pi n}{3}-1\right)}{\pi^{2}n^{2}}\cos\dfrac{2n\pi}{3}x\right)$.

\subsection{$f(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x<0, \\ x^{2}, & 0 \leqslant x<1 ;\end{array}\right.$}
\textbf{解}\quad
将$f(x)$延拓,其周期为$2T=2$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\mathrm{~d}x                       \\
		      & =\dfrac{1}{T}\int_{0}^{T}x^{2}\mathrm{~d}x=\dfrac{1}{3}          \\
		a_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\cos \dfrac{n\pi}{T} x\mathrm{~d}x \\
		      & =\int_{0}^{1}x^{2}\cos n\pi x\mathrm{~d}x                        \\
		      & =\dfrac{(-1)^{n}2}{n^{2}\pi^{2}}                                 \\
		b_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\sin \dfrac{n\pi}{T} x\mathrm{~d}x \\
		      & =\int_{0}^{1}x^{2}\sin n\pi x\mathrm{~d}x                        \\
		      & =\dfrac{2((-1)^{n}-1)-(-1)^{n}n^{2}\pi^{2}}{n^{3}\pi^{3}}
	\end{aligned}
$$
由于$f(x)$分段可微,从而在$(-1,1)$中,
$$
	f(x)=\dfrac{1}{6}+\sum_{n=1}^{\infty}\left(\dfrac{(-1)^{n}2}{n^{2}\pi^{2}}\cos n\pi x+\dfrac{2((-1)^{n}-1)-(-1)^{n}n^{2}\pi^{2}}{n^{3}\pi^{3}}\sin n\pi x\right)
$$
在间断点处级数收敛到$\dfrac{f(x--0)+f(x+0)}{2}=\dfrac{1}{2}$.
\subsection{$f(x)=\left\{\begin{array}{ll}2 x+1, & -3 \leqslant x<0, \\ 1, & 0 \leqslant x<3 .\end{array}\right.$}
\textbf{解}\quad
将$f(x)$延拓,可知$f(x)$周期为$2T=6$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\mathrm{~d}x                                                                                   \\
		      & =\dfrac{1}{3}\int_{-3}^{3}f(x)\mathrm{~d}x                                                                                   \\
		      & =\dfrac{1}{3}\left(\int_{-3}^{0}(2x+1)\mathrm{~d}x+\int_{0}^{3}1\mathrm{~d}x\right)=-1                                       \\
		a_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\cos \dfrac{n\pi}{T} x\mathrm{~d}x                                                             \\
		      & =\dfrac{1}{3}\left(\int_{-3}^{0}(2x+1)\cos \dfrac{n\pi}{3}x\mathrm{~d}x+\int_{0}^{3}\cos \dfrac{n\pi}{3}x\mathrm{~d}x\right) \\
		      & =\dfrac{6(1-(-1)^{n})}{\pi^{2}n^{2}}                                                                                         \\
		b_{n} & =\dfrac{1}{T}\int_{-T}^{T}f(x)\sin \dfrac{n\pi}{T} x\mathrm{~d}x                                                             \\
		      & =\dfrac{1}{3}\left(\int_{-3}^{0}(2x+1)\sin\dfrac{n\pi}{3}x\mathrm{~d}x+\int_{0}^{3}\sin\dfrac{n\pi}{3}x\mathrm{~d}x\right)   \\
		      & =\dfrac{6(-1)^{n}}{\pi n}
	\end{aligned}
$$
所以由于$f(x)$分段可微,在$[-3,3)$上有
$$
	f(x)=-\dfrac{1}{2}+\sum\limits_{n=1}^{\infty}\left(\dfrac{6(1-(-1)^{n})}{\pi^{2}n^{2}}\cos\dfrac{n\pi}{3}x+\dfrac{6(-1)^{n}}{\pi n}\sin\dfrac{n\pi}{3}x\right)
$$
在间断点处级数收敛到$-2$.
\section{}
\subsection{在 $(-\pi, \pi)$ 上展开 $|\sin x|$ 为 Fourier 级数, 并证明 $\dfrac{1}{2}=\sum\limits_{n=1}^{\infty} \dfrac{1}{4 n^{2}-1}$}
\textbf{解}\quad
$|\sin x|$是偶函数,从而$b_{n}=0$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}|\sin x|\mathrm{~d}x                        \\
		      & =\dfrac{2}{\pi}\int_{0}^{\pi}\sin x\mathrm{~d}x=\dfrac{4}{\pi}                                                             \\
		a_{n} & =\dfrac{2}{\pi}\int_{0}^{\pi}\sin x \cos nx\mathrm{~d}x                     \\
		      & =\dfrac{1}{\pi}\left(\int_{0}^{\pi}\sin(1-n)x+\sin(1+n)x\mathrm{~d}x\right) \\
		      & =-\dfrac{2(1+(-1)^{n})}{\pi(n^{2}-1)}
	\end{aligned}
$$
所以由$f(x)$分段可微,没有间断点,所以
$$
	f(x)=\dfrac{2}{\pi}-\sum_{n=1}^{\infty}\left(\dfrac{2(1+(-1)^{n})}{\pi(n^{2}-1)}\cos nx\right)
$$
令$x=0$得
$$
	\begin{aligned}
		\dfrac{2}{\pi} & =\sum_{n=1}^{\infty}\left(\dfrac{2(1+(-1)^{n})}{\pi(n^{2}-1)}\cos nx\right) \\
		               & =\sum_{n=1}^{\infty}\dfrac{4}{\pi(4n^{2}-1)}
	\end{aligned}
$$
所以
$$
	\dfrac{1}{2}=\sum\limits_{n=1}^{\infty} \dfrac{1}{4 n^{2}-1}
$$

\subsection{在 $(-\pi, \pi)$ 上展开 $|x|$ 为 Fourier 级数,并证明 $\dfrac{\pi^{2}}{8}=\sum\limits_{n=1}^{\infty} \dfrac{1}{(2 n-1)^{2}}$}
\textbf{解}\quad
$|x|$是偶函数,所以$b_{n}=0$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}|x|\mathrm{~d}x=\pi                                                   \\
		a_{n} & =\dfrac{1}{\pi}\int_{-\pi}^{\pi}|x|\cos nx\mathrm{~d}x \\
		      & =\dfrac{2}{\pi}\int_{0}^{\pi}x\cos nx\mathrm{~d}x      \\
		      & =\dfrac{2((-1)^{n}-1)}{n^{2}\pi}
	\end{aligned}
$$
从而由于$f(x)$分段可微且无间断点,所以
$$
	\begin{aligned}
		f(x) & =\dfrac{\pi}{2}+\sum_{n=1}^{\infty}\left(\dfrac{2((-1)^{n}-1)}{n^{2}\pi}\cos nx\right) \\
		     & =\dfrac{\pi}{2}-\dfrac{4}{\pi}\sum_{n=1}^{\infty}\dfrac{1}{(2n-1)^{2}}\cos(2n-1)x
	\end{aligned}
$$
取$x=0$即得$\dfrac{\pi^{2}}{8}=\sum\limits_{n=1}^{\infty}\dfrac{1}{(2n+1)^{2}}$.

\subsection{在 $(0,2 \pi)$ 上展开 $f(x)=\dfrac{1}{12}\left(3 x^{2}-6 \pi x+2 \pi^{2}\right)$ 的 Fourier 级数,并证明 $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6}$}
\textbf{解}\quad
将其延拓,$f(x)$是偶函数且没有间断点,且其分段可微.
$$
	\begin{aligned}
		a_{0} & =\dfrac{1}{\pi}\int_{0}^{2\pi}f(x)\mathrm{~d}x=0                                       \\
		a_{n} & =\dfrac{1}{12\pi}\int_{0}^{2\pi}\left(3x^{2}-6\pi x+2\pi^{2}\right)\cos nx\mathrm{~d}x=\dfrac{1}{n^{2}}
	\end{aligned}
$$
所以
$$
	f(x)=\sum_{n=1}^{\infty}\dfrac{1}{n^{2}}\cos nx
$$
令$x=0$即可得$\dfrac{\pi^{2}}{6}=\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{2}}$.

\section{设 $\alpha$ 不是整数, $f(x)=\cos \alpha x, \quad x \in[-\pi, \pi] .$}
\subsection{将 $f(x)$ 展开成 Fourier 级数}
\textbf{解}\quad
$f(x)$是偶函数,所以$b_{n}=0$.
$$
	\begin{aligned}
		a_{0} & =\dfrac{2}{\pi}\int_{0}^{\pi}\cos \alpha x\mathrm{~d}x                                \\
		      & =\dfrac{2}{\pi\alpha}\sin \alpha x                                                    \\
		a_{n} & =\dfrac{2}{\pi}\int_{0}^{\pi}\cos \alpha x \cos nx\mathrm{~d}x                        \\
		      & =\dfrac{1}{\pi}\int_{0}^{\pi}\left[\cos(\alpha-n)x+\cos(\alpha+n)x\right]\mathrm{~d}x \\
		      & =\dfrac{(-1)^{n}2\alpha\sin(\alpha\pi)}{\pi(\alpha^{2}-n^{2})}
	\end{aligned}
$$
所以
\begin{equation}
	f(x)=\dfrac{\sin\alpha \pi}{\pi\alpha}+\sum_{n=1}^{\infty}\dfrac{(-1)^{n}2\alpha\sin(\alpha\pi)}{\pi(\alpha^{2}-n^{2})}\cos nx
\end{equation}

\subsection{证明: $ \dfrac{\pi}{\sin a \pi}=\dfrac{1}{a}+\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{2 a}{a^{2}-n^{2}}$}
\begin{proof}
	对式(1)取$x=0$,两边同乘$\dfrac{\pi}{\sin\alpha\pi}$即可得
	\begin{equation}
		\dfrac{\pi}{\sin a \pi}=\dfrac{1}{a}+\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{2 a}{a^{2}-n^{2}}
	\end{equation}
\end{proof}


\subsection{证明: $\dfrac{1}{\sin x}=\dfrac{1}{x}+\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{2 x}{x^{2}-n^{2} \pi^{2}}, \quad 0<x<\pi$}
\begin{proof}
	对(2)两边同除$\pi$,并令$\alpha\pi=x$即可得$\dfrac{1}{\sin x}=\dfrac{1}{x}+\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{2 x}{x^{2}-n^{2} \pi^{2}}, \quad 0<x<\pi$
\end{proof}




\end{document}